Fig. 4. The optimal integer solution inside of the possible set

Ŕ periodica polytechnica

Social and Management Sciences 19/2 (2011) 87–96 doi: 10.3311/pp.so.2011-2.05 web: http://www.pp.bme.hu/so c Periodica Polytechnica 2011 RESEARCH ARTICLE

An alternative solution to an economic order quantities for recoverable item inventory systems

ImreDobos

Received 2011-06-07

Abstract

In this paper we analyze a reverse logistics inventory model investigated by Teunter (2001). Demand of a product can be sat- isfied with newly manufactured and remanufactured products.

The used products are collected and then returned to the man- ufacturer of the product. The returned items are inspected and some of the used products are disposed outside. The rate of reuse is decision variable in this model. The goal of the decision maker is to minimize the relevant costs. The aim of this paper is to reexamine the proposed model and to improve some small mistakes. First, we show that it is comfortable to determine the batch numbers of the manufacturing and remanufacturing activ- ities, because these decision variables are integer, rather than the manufacturing and remanufacturing lot sizes. Secondly, we show that there exist optimal solutions for the batch numbers inside of the possible set, i.e. both manufacturing and remanu- facturing batches could be higher than one. Thirdly, the above mentioned paper have not analyzed a case with one batch num- bers both manufacturing and remanufacturing. This case can occur for some parameters.

Keywords

Reverse Logistics · Closed-loop supply chain · Lotsizing · EOQ·Optimization

Acknowledgement

The author gratefully acknowledges the financial sup- ports from the TÁMOP-4.2.1.B-09/KMR- 2010-0005 research program and the Deutscher Akademischer Austauschdienst (DAAD).

Imre Dobos

Institute of Business Economics, Corvinus University Budapest, H-1093 Bu- dapest, F˝ovám tér 8., Hungary

e-mail: imre.dobos@uni-corvinus.hu

1 Introduction

Quantitative models for inventory systems with product re- covery management provide an actual generalization of clas- sical EOQ models. The classical EOQ model analyzes one product inventory systems. The difficulty of recovery system is that manufactured (purchased) items must be handled simul- taneously in a manufacturing-remanufacturing cycle. A number of authors have proposed such models, like Schrady [4], Nah- mias and Rivera [2], and Richter [3]. Our paper deals with one of these proposals, we investigate the model of Teunter [5].

The goal of the paper is to reconsider the Teunter‘s model.

First, the explicit model will be discussed and a way of solution is offered for this model, because the author has neglected to describe the explicit model for a general inventory holding pol- icy, i.e. with explicit manufaturing and remanufacturing batch numbers. The model is solved in two ways. In the first method we eliminate the batch numbers, in order to calculate the manu- facturing and remanufacturing lot sizes. This way of solution is common in the EOQ-type inventory literature, but it makes dif- ficult to determine the integer batch numbers for manufacturing and remanufacturing. The second method offers to eliminate the lot sizes, and then to determine the batch numbers. This way leads to a meta-model offered by Richter [3], and analyzed mathematically by Dobos and Richter [1]. This model makes it easier to calculate the optimal reuse rate of the model.

The paper organizes, as follows. In the second section the model is shown and the cost function of the problem is con- structed. Section three presents the possible ways to solve the problem, and shows the integer solution of the model. The next part deals with the determination of the optimal reuse rate, and last we sumatize the results of the paper.

2 The model and the cost function

Teunter [5] has investigated in his model the following activ- ities:

– remanufacturing, – disposal and – manufacturing.

Let a cyclebe deined, as the above-mentioned schedule of activities with fixed batch sizes for manufacturing and reman- ufacturing. In a planning period there isonly onecycle. (This can be proved very easily by grouping the remanufacturing and manufacturing lots.)

The goal of the decision maker is to minimize the relevant costs for manufacturing and remanufacturing. There are EOQ- oriented setup and holding costs for remanufacturing and manu- facturing, and linear production and remanufacturing costs, lin- ear disposal cost and holding cost for non-serviceable items.

The notations of the model are the following:

System parameters:

r return rate (0≤r≤1), λ rate of demand.

Cost parameters:

Km setup cost for manufacturing, Kr setup cost for remanufacturing, hm holding cost for manufactured items, hr holding cost for remanufactured items, h_n holding cost for non-serviceable items, c_m manufacturing cost,

c_r remanufacturing cost,

c_d cost for disposing one non-serviceable item.

Decision variables:

Q_m batch size for manufacturing, Q_r batch size for remanufacturing,

M number of manufacturing batches, positive inte- ger,

R number of remanufacturing batches, positive inte- ger,

T length of the product recovery cycle, u reuse rate (0≤u≤r).

We assume that all parameters and the decisions variables are nonnegative numbers. We will describe the mathematical model with some application.

First we examine stock-flow balance of serviceable and re- coverable stocks. Eq. (1) shows that the sum of manufactured and remanufacted products must cover the demand in a cycle.

Eq. (2) is the relation between the returned products and the use of these products for remanufacturing and disposal. The mate- rial flow of the model is shown in Fig. 1.

M Q_m+R Q_r =λT (1)

R Qr+(r−u)λT =r T (2)

From the linear systems (1) and (2) we can write two separate equations for the manufacturing and remanufacturing batches:

M Q_m =(1−u)λT (3) and

R Qr =uλT (4)

If the reuse rate is equal to zero, i.e.u =0, then the remanufac- turing lot size is zero, i.e. Qr =0in relation (4). It means that all returned parts are disposed, there is no reuse in system and the management problem turns into a simple inventory problem.

Another interesting case is, if the return rate is equal to reuse rate (u =r). This case shows an example, when all returned parts are reused and there is no disposal activity. Identity (3) and (4) will be useful to create our cost function.

Now we construct the total cost function. We do it in two steps. In the first step we investigate the inventory holding cost function H(Q_m,Q_r,T,M,R,u)for serviceable and non- serviceable parts. In the second step we describe the linear costs L(Qm,Qr,T,M,R,u)of manufacturing, remanufactur- ing, and disposal.

Let us now calculate the inventory holding costsH(Qm,Qr, T, M, R, u). The inventory holding policy is shown in Fig. 2. This inventory holding policy presents a manufac- turing/remanufacturing strategy, i.e. a cycle. A manufactur- ing/remanufacturing cycle is an order of these activities. In Fig. 2 we have shown a cycle beginning with given remanu- facturing batches and then followed with given manufacturing batches. As in the classical EOQ model, these fixed cycles are repeated infinitely. Of course, we can change the order of the ac- tivites, i.e. a cycle can begin with manufacturing and followed with remanufacturing, but this order does not change the inven- tory holding costs. This policy is a predetermined policy and we look for the optimal parameters (Qm,Qr,T,M,R,u)of this strategy. Let us assume that the inventory level functions for a known strategy are function I^s(t)for serviceable stock and function I^r(t)for recoverable stock, 0≤ t ≤T. The inven- tory holding costs are the area below this functions, i.e.

H(Q_m,Q_r,M,R,u)=

h_r

T u

Z

0

I^s(t)dt+h_m

T

Z

T u

I^s(t)dt+h_n

T

Z

0

I^r(t)dt.

Now we use the property of the inventory policy that the sum of serviceable and recoverable products is a monotone decreas- ing, linear and continuous function of time in the remanufac- turing cycle. So the inventory cycle can be divided into two subcycles.

1 the demand is satisfied from remanufacturing, and the recov- erable stock is positive. The length of this interval is equal to T u−^Q_λ^r.

2 The demand is satisfied from the last remanufacturing batch and from manufacturing, and the stock level of recoverable items is monotone nondecreasing. A remanufacturing batch is used in an interval length of ^Q_λ^r. The length of this subcycle isT(1−u)+^Q_λ^r.

So the inventory holding cost function can be expressed with

Fig. 1. Material flow in the model

Market

Disposal Remanufacturing

Used products

Final products M⋅Q_m =(1-u)⋅λ⋅T

R⋅ Q_r =u⋅λ⋅T

R⋅ Q_r =u⋅λ⋅T

λ⋅T

r⋅λ⋅T

(r-u)⋅λ⋅T Manufacturing

From the linear systems (1) and (2) we can write two separate equations for the manufacturing and remanufacturing batches:

Fig. 1. Material flow in the model

Fig. 2. Modeling the inventory policy (R=3, M=7)

Recoverable stock I^r(t) Serviceable stock I^s(t)

T·u

Remanufacturing Manufacturing

Q_r

rλ

r T u r− λ^r Q

T -λ

Q_r Q_m

T·(1-u)

T

Fig. 2. Modeling the inventory policy (R=3,M=7)

the help of the cycles, as H(Q_m,Q_r,M,R,u)= (h_r−h_n)

T u−^Qr_λ

R

0

I^s(t)dt+h_r

T u

R

T u−^Qr_λ

I^s(t)dt+h_m

T

R

T u

I^s(t)dt+

h_n

T u−^Qr_λ

R

0

I^s(t)+I^r(t) dt+h_n

T

R

T u−^Qr_λ

I^r(t)dt .

We must now calculate the five integrals. The first integral con- sists of R-1 pieces of remanufacturing batches. The costs are (h_r −h_n)

T u−^Qr_λ

R

0

I^s(t)dt =(R−1) (h_r −h_n)^Q_2λ^r2. The second integral is only a remanufacturing batch h_r

T u

R

T u−^Qr_λ

I^s(t)dt =

hrQ²_r

2λ. The third value is the cost of inventory holding of manu-

factured products, which consists ofMbatchesh_m

T

R

T u

I^s(t)dt = M hmQ²_m

2λ. The computation of the fourth integral is a little bit complicated. We have pointed out that the sum of the inventory levels I^s(t)+I^r(t)is a monotone decreasing linear function.

The tangent of this linear function is (1-r)λ. In point of time T u− ^Q_λ^r this function has a value of Q_r. With this assump- tion the value of the integral ishn

T u−^Qr_λ

R

0

I^s(t)+I^r(t) dt = h_nh

Q_r +(1−r)^λ₂

T u− ^Q_λ^ri

T u− ^Q_λ^r

. The fifth, last in- tegral ish_n

T

R

T u−^Qr_λ

I^r(t)dt=h_nr^λ₂

1−r

r T u+ ^Q_λ^r2

.

Summing up the integrals and with elementary calculations, we have the following expression for the inventory holding costs:

H(Q_m,Q_r,T,M,R,u)= Rh_rQ²_r

2λ +M h_mQ²_m

2λ +h_nQ²_r 2λ

R²1−r r +R

. In this expression we have applied from Eq. (4) thatT u= ^{R Q}_λ^r.

The linear costs of manufacturing, remanufacturing, and dis- posal can be calculated very easily:

L(Q_m,Q_r,T,M,R,u)=c_mMQ_m+c_rRQ_r +c_dλT(r-u)= Tλ[u(cr −cm−cd)+(cm+cdr)]

Now we can formulate the average cost function Ca(Qm,,Qr,T,M,R,u)summing up the total inventory holding costs (setup and inventory holding costs) and the linear man- ufacturing, remanufacturing, and disposal costs, and divided with the length of the cycle:

Ca(Qm,Qr,T,M,R,u)=

= ^{R Kr+Rhr}

Q2

2λr+M Km+M hmQ2 2λm+hnQ2

2λr

n

R^{2 1−r}_r +Ro

T +

+uλ (cr−cm−cd)+λ (cm+cdr)

This way we have constructed a non-linear mixed-integer math- ematical programming problem:

Ca(Qm,QrT,M,R,u)→min such that

M Qm =(1−u)λT, R Qr =uλT, 0≤u≤r,

Q_m≥0, Q_r ≥0, T >0, M,Rpositive integers





















 (P)

The problem (P) consists of two subproblems. The first prob- lem is to determine the optimal inventory related decision vari- ables (Q_m,Q_r,T,M,R), and the second problem is to calculate the optimal reuse rateu. Of course, we can change the order of the manipulation beginning with the optimal reuse rate. For the sake of simplicity we have decided to determine first the in- ventory related variables. In the next section we will solve the problem for the relevant variables.

3 Determination of the optimal inventory related deci- sion variables

In this section we solve problem (P) in two different way. The difference is the order of eliminating the continuous variables from the cost function using Eqs. (3) and (4). The first method offers to eliminate the integer variables R and M, in order to express the manufacturing and remanufacturing lot sizes. Sec- ond method suggests elimination of the lot sizes, in order to in- vestigate an integer programming problem. The reuse rateu is parameter in these problems. Let us follow this two ways.

3.1 Elimination of batch numbersRandM

In this case the author will express the cost function in de- pendence on the lot sizes Qm andQr, so he must substitute the number of batches in the goal function, i.e. M = ⁽¹⁻_Q^u_m^)λT, and R= ^u_Q^λ_r^T, which are not smaller than one. With the help of this substitution we can reformulate the problem (P) into the prob- lem (P^T)as

u_λ

Kr

Qr +^hr+₂^hnQ_r + (1−u)_λ

Km

Qm +^h₂^mQ_m +

hn

2u²λT

1 r −1

+

+uλ (c_r−c_m−c_d)+λ (c_m+c_dr)→min such that

(1−u)λT Qm ≥1,

uλT Qr ≥1,

Q_m≥0, Q_r ≥0, T >0.









































 (P^T)

This new cost function is linear in the length of the cycle, and for this length there is a lower bound. It means that, T = ^Q_uλ^r and/orT = ₍₁^Q₋^m_u)λ. If the first inequality is equation, i.e.R^o=1, then the cost function is

u λKr

Q_r +^hr+

1 rh_n 2 Qr

+(1−u) _λ

Km

Qm +^h₂^mQ_m +

+uλ (c_r−c_m−c_d)+λ (c_m+c_dr)

The optimal lot sizes are in this case Q^o_r =

r 2λKr

hr+¹_rhn, and Q^o_m =

q2λK_m

h_m . From these equalities it follows that ⁽¹⁻_Q^u_o^)λT

m ≥

uλT

Q_r^o =1, i.e.(1−u)Q^o_r ≥u Q^o_m. This asssumption is held after substitution the optimal lot sizes in the inequality, if

0≤u ≤min











√Krhm

√Krhm+ r

Km

hr+¹_rhn

; r









 .

The cost function is now after substitution the lot sizes in the cost function

u r

2λK_r

h_r+¹_rh_n

+ (1 − u)√

2λK_mh_m +

+uλ (cr−cm−cd) + λ (cm+cdr) . The optimal cycle time is nowT^o=_u¹

r2 λ Kr

h_r+¹_rh_n.

If the second inequality is an equation, i.e. M^o =1, then the cost function is

u_λ

Kr

Qr +^hr+₂^hnQ_r

+(1−u)·

λKm

Qm +

h_m+_(1−u)^u2₂

1 r−1

h_n

2 Q_m

! + +uλ (c_r −c_m−c_d)+λ (c_m+c_dr) The optimal lot sizes for this second model areQ_r^o=

q2λK_r h_r+h_n, and Q^o_m = (1 − u)r _2λK

m

(1−u)²hm+u² 1

r−1

hn

. From the first two inequalities of the problem (P^T) it follows that in this case ^uλT_Qo

r ≥ ⁽¹⁻_Q^uo^)λT

m = 1, i.e. u Q^o_m ≥ (1 − u)Q^o_r. This last assumption can be reformulated with the help of optimal lot sizes after substitution and some calculations as

√K_rh_m

√K_rh_m+ r

K_m(h_r+h_n)−K_rh_n

1

r−1 ≤u≤r. This interval foruex- ists, if the inequality

√Krhm

√K_rh_m+ r

K_m(h_r+h_n)−K_rh_n

1

r−1 ≤rholds.

We can calculate this smallest value ofu.

The cost function has the next form after the substitution the optimal lot sizes:

u√

2λK_r(h_r +h_n)+

r 2λKm

h(1−u)²hm+u²

1 r −1

hn

i + +uλ (c_r−c_m−c_d)+λ (c_m+c_dr)

.

This model depends only on the reuse rateu. The optimal length of a cycle is

T^o= v u u t 2 λ

K_m (1−u)²h_m+u²

1 r −1

h_n .

These two cost functions and lot sizes for manufacturing and remanufacturing are the same as those of Teunter [4]. We have formulated necessary conditions, when to use these models in dependence on the reuse rateu. We see that there is no lot sizes for manufacturing and remanufacturing, if

min











√K_rh_m

√K_rh_m+ r

K_m

h_r+h_n¹_r, r











≤u≤

√K_rh_m

√Krhm+ r

Km(hr+hn)−Krhn

1 r −1

.

The question is how to solve the model on this interval of the reuse rate. To answer these questions we must generalize the offered model with the case, when both numbers of batches for manufacturing and remanufacturing are equal to one. We will see that this case is investigated in the model suggested by Richter [3], as well.

The not investigated case is M^o = R^o =1. For this case T = ₍₁^Q₋^m_u)λ = ^Q_uλ^r, and ^u_Q^λT

r = ⁽¹⁻_Q^u_m^)λT = 1. Let us know substitute expressionsT = ₍₁^Q₋^m_u)λ, and Qr = ₁₋^u_uQm in the cost function. Then we have

(1−u)^λ(K_Q^r+_m^Km) +₁₋¹_u¹₂· h

h_r+¹_rh_n

u²+h_m(1−u)²i Q_m+ +uλ (cr −cm−cd)+λ (cm+cdr)

The optimal lot sizes are for this third case Q^o_m = (1 − u)r

2λ(Kr+Km) h

hr+¹_rhn

u²+hm(1−u)²i, and Q^o_r = ur

2λ(Kr+Km) h

hr+¹_rhn

u²+hm(1−u)²i. This type of model is valid,

if √

Krhm

√K_rh_m+ r

K_m

h_r+h_n¹_r ≤u ≤

√Krhm

√K_rh_m+ r

K_m(h_r +h_n)−K_rh_n

1 r −1. The modified cost function is

s

2λ (K_r+K_m)

h_r +1 rh_n

u²+h_m(1−u)²

+

uλ (cr −cm−cd)+λ (cm+cdr) , and the optimal length of a cycle is

T^o= v u u t 2 λ

K_r+K_m h

hr +¹_rhn

u²+hm(1−u)²i.

The optimal continuous solution of this method for lot sizes, number of batches and cycle time is in dependence on the reuse rateu.

Theorem 1 The optimal continuous solution for the inventory related costs and cost function are

1 If 0≤u≤min







√K_rh_m

√Krhm+ r

Km

hr+¹_rhn

; r





 , then

Q^o_r =

s 2λK_r h_r +¹_rh_n,

Q^o_m = s

2λK_m

h_m , T^o= 1 u

s2 λ

K_r h_r +¹_rh_n, R(u)=1,M(u)= ¹⁻_u^u

r Krhm

Km(hr+hn)+Kmhn1−r r

,

C^T(u)=u s

2λK_r

h_r+1 rh_n

+

(1−u)p

2λK_mh_m+uλ (c_r −c_m−c_d)+λ (c_m+c_dr) .

2 If

√Krhm

√Krhm+ r

Km

hr+hn·¹r

≤u≤

√Krhm

√Krhm+ r

Km(hr+hn)−Krhn

1 r−1

, then, Q^o_m =(1−u)r _2λ(K

r+Km) h

hr+¹rhn

u²+hm(1−u)²i, Q^o_r =ur _2λ(K

r+Km) h

hr+¹rhn

u²+hm(1−u)²i, T^o= r

2λ Kr+Km

h hr+¹rhn

u²+hm(1−u)²i,R(u)=1,M(u)=1, C^T(u)=

s

2λ (Kr +Km)

hr+1 rhn

u²+hm(1−u)² +uλ (c_r −c_m−c_d)+λ (c_m+c_dr) . 3 If

√Krhm

√Krhm+ r

Km(hr+hn)−Krhn

1 r−1

≤u ≤r, then

Q^o_r = s

2λK_r h_r+h_n,

Q^o_m =(1−u) v u u t

2λK_m (1−u)²hm+u²

1 r −1

hn

, R(u)=u

r Km(hr+hn) Krhm(1−u)²+Krhn1−r

r u² M(u)=1, C^T(u)=up

2λK_r(h_r +h_n)+

s 2λK_m

(1−u)²h_m+u² 1

r −1

h_n

+ uλ (cr −cm−cd)+λ (cm+cdr) .

Remark. In the theorem 1 we can see, if the reuse rate is rel- atively small, then only one remanufacturing batch occurs in a recovery cycle. Nevertheless, if the reuse rate is relatively high, then only one manufacturing batch is used in the cycle and the rest of the demand is satisfied from remanufactured products.

This result can be relevant for practical application. If the col- lection rate of used products is low, then it is enough to plan inly one remanufacturing batch in a planning period, and inversely.

If the collection rate and the reuse rate is high, than the demand for the new items can be satisfied from remanufactured products and the lacked amount of items to manufacture.

Let us now analyze the optimal solution in dependence of the reuse rateu and return rater. The possible sets of parameters of the three cases are examined parametrically. Let us define the following two functions:

u₁(r)=

√Krhm

√K_rh_m+ r

K_m

h_r+h_n¹_r, and

u2(r)=

√K_rh_m

√Krhm+ r

Km(hr +hn)−Krhn

1 r −1.

Fig. 3. The possible set of parameters u and r

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

u1 r( ) u2 r( ) r

r2 r

r₁

Fig. 3. The possible set of parametersuandr

It is easy to see that function u₁(r) is smaller than function u2(r), butu1(1)=u2(1).

If we draw these two functions, then we get Fig. 3.

Fig. 3 shows pointsr1andr2. These two points are the switch- ing points between the solution regions. If the known return rate ris smaller than valuer1, then only case (1) of theorem 1 occurs in the solution of the problem (P^T). If reuse rateris betweenr1

andr₂, then cases 1 and 2 occur only. If reuse rateu is greater thanr₂, then all three cases are in the optimal solution. The cost functionC^T(u)is continuous at pointsr₁andr₂. The dotted vertical line of Fig. 3 shows a general case, when all three cases occur in the solution foru.

The continuous solution of problem (P^T)is known with these last expressions. The question is now, how to determine the integer batch numbers for manufacturing and remanufacturing.

This solution method does not give algorithm to determine the integer solution for problem (P^T).

3.2 Elimination of lot sizes Q_r and Q_m

We can follow an other way to solve the problem. Let us substitute the manufacturing and remanufacturing lot sizes Q_m = ⁽¹⁻_M^u)λT and Q_r = ^uλ_R^T in the cost function from Eqs.

(3) and (4). After substitution the problem has the next form:

R Kr+M Km

T +

T hhr+hn

2 u²λ¹_R+^h₂^m(1−u)²λ_M¹ +^h₂ⁿu²λ 1 r−1

i +uλ (c_r−c_m−c_d)+λ (c_m+c_dr)→min

such that T>0,

M,Rpositive integers.





















 (P^R)

The cost function is now convex in the length of the cycle, so the optimal length can be calculated as follows:

T^o(M,R,u)=

v u u t

R Kr+M Km hr+hn

2 u²λ_R¹ +^h₂^m(1−u)²λ_M¹ +^h₂ⁿu²λ

1 r −1

. After substitution the optimal length in the cost function, we have the following cost functionCI(R,M,u):

C_I(M,R,u)= r

(R K_r+M K_m)h_h

r+hn

2 u²λ_R¹+^h₂^m(1−u)²λ_M¹ +^h₂ⁿu²λ 1 r −1i

+ +uλ (c_r−c_m−c_d)+λ (c_m+c_dr)

The functionCI(R,M,u)can be written in the next form C_I(M,R,u)=

q

2λ A(u)_M^R +B(u)^M_R +C(u)R+D(u)M+E(u) + λu(cr−cm−cd)+λ (cm+cdr)

(5) where

A(u)=Krhm(1−u)², B(u)=Km(hr +hn)u², C(u)=K_rh_n

1 r −1

u², D(u)=K_mh_n

1 r −1

u²,

E(u)=K_r(h_r+h_n)u²+K_mh_m(1−u)²

,

and variablesRandMare positive integers, and0≤u≤r.

The manufacturing and remanufacturing lots are in this case Q_m(M,R,u)= (1−u)λ

M . v

u u t

R K_r +M K_m

hr+hn

2 u²λ_R¹ +^h₂^m(1−u)²λ_M¹ +^h₂ⁿu²λ

1 r −1 and

Q_r(M,R,u)= uλ R. v

u u t

R K_r+M K_m

h_r+h_n

2 u²λ_R¹ +^h₂^m(1−u)²λ_M¹ +^h₂ⁿu²λ

1 r −1. FunctionC_I(R,M,u)is quasiconvex in RandM and convex inu. This property guarantees the existence of optimal solution, as it is prooved by Dobos and Richter [5]. Let us now introduce an auxilliary functionS(R,M,u), as follows

S(R,M,u)= A(u)R

M +B(u)M R+ C(u)R+D(u)M+E(u).

We look for an optimal solution of this function for remanufac- turing and manufacturing batchesRandM. This function is the expression under the square in (1). Due to monotonicity con- siderations the function S(R,M,u)can be analysed for solving batch sizes R and M, where all coefficients A(u), B(u), C(u), D(u)andE(u)are positive. The used meta-model is shown in the appendix of this paper. After some calculation we get the possible set of reuse rateu.

Theorem 2 The continuous optimal solution for minimizing the functionCI(R,M,u)forRandMis

(i)u ≤min







√K_rh_m

√Krhm+ r

Km

hr+hn1

r

;r





 R(u)=1, M(u)=¹⁻_u^ur

Krhm

K_m(h_r+h_n)+K_mh_n^1−r_r ,

C^R(u)=u s

2λKr

hr+1

rhn

+(1−u)p

2λKmhm+

uλ (c_r −c_m−c_d)+λ (c_m+c_dr) , (ii)

√K_rh_m

√K_rh_m+ r

K_m

h_r+h_n¹_r ≤u≤

√Krhm

√Krhm+ r

Km(hr+hn)−Krhn

1 r−1

R(u)=1,M(u)=1, C^T(u)=

s

2λ (K_r+K_m) h_r +1

rh_n

u²+h_m(1−u)² + uλ (c_r −c_m−c_d)+λ (c_m+c_dr) ,

(iii)

√Krhm

√Krhm+ r

Km(hr+hn)−Krhn

1

r−1 ≤u ≤r R(u)=u

r Km(hr+hn)

K_rh_m(1−u)²+K_rh_n^1−r_r u², M(u)=1, C^T (u)=up

2λK_r(h_r+h_n)+

s 2λK_m

(1−u)²h_m+u² 1

r −1

h_n

+ uλ (c_r −c_m−c_d)+λ (c_m+c_dr) .

As we see, the optimal continuous solutions are the same as be- fore. But this formulation of the problem makes it possible to investigate the integer solution of the model (1), as it is made in paper Dobos and Richter [1].

In the next section we outline the integer solution of model (1).

3.3 Integer solution of model (1)

The optimal continuous solution of problem (P) is on the bor- der of the possible sets of (R(u),M((u)), i.e. at least one of the batch is equal to one. The question is whether the optimal integer solution is on the border, or there is optimal integer so- lution inside of the possible set, i.e. both of the batch numbers are strictly greater than one. To answer the question, we first determine the integer solution on the border of the possible set.

Theorem 3 (Dobos and Richter[1])

The integer optimal solution for minimizing the function C_I(R,M,u)for RandM is

1 u≤min







√K_rh_m

√K_rh_m+ r

K_m

h_r+h_n¹_r;r





 R^g(u)=1,

M^g(u)=

$ r

1−u u

2

K_rh_m Km(hr+hn)+Kmhn1−r

r

+¹₄

% +¹₂,

2

√Krhm

√Krhm+ r

Km

hr+hn1

r

≤u≤

√Krhm

√Krhm+ r

Km(hr+hn)−Krhn

1 r−1

R^g(u)=1,M^g(u)=1, 3

√Krhm

√K_rh_m+ r

K_m(h_r+h_n)−K_rh_n

1

r−1 ≤u ≤r R^g(u)=

r

u^{2 Km(hr+hn)}

Krhm(1−u)²+Krhn1−r r u² +¹₄

+¹₂, M^g(u)=1,

Notation|x|denotes the greatest integer not higher than valuex.

We will not prove this theorem, the proof is can be seen in paper of Dobos and Richter [5]. The border batch numbers (R^g(u), M^g(u))are optimal for some cases, but there exist optimal so- lutions inside of the possible set of batch numbers. In the next section we show some example to demonstrate the reults of the theorem.

It can be proven that the inventory related cost difference be- tween the a border solution (R^g(u),M^g(u))and an optimal so- lution inside of the possible set is not higher than that of 2.1 percent (Dobos and Richter [1]). The optimal integer solution can be calculated, as

Q_m M^g(u) ,R^g(u) ,u

= (1−u)λ M

v u u t

R^g(u)K_r+M^g(u)K_m hr+hn

2 u²λ_Rg¹(u)+^h₂^m(1−u)²λ_Mg¹(u)+^h₂ⁿu²λ 1 r −1

,

and

Q_r M^g(u) ,R^g(u) ,u

=uλ R

v u u t

R^g(u)K_r+M^g(u)K_m hr+hn

2 u²λ_Rg¹(u)+^h₂^m(1−u)²λ_Mg¹(u)+^h₂ⁿu²λ 1 r −1.

3.4 Numerical examples Example 1.

Let the parameters be in the next example:

λ=1.000 piece/year,r=0.9,u=0.5, Km =$ 750,Kr =$ 100,

hm=$ 200,hr =$ 50, hn=$ 20,

c_m =$ 20,c_r =$ 15,c_d=$ -35.

In this example the remanufacturing firm sells the disposed items. The optimal continuous solution for this problem is:

R^o=1.611,M^o=1, Q^o_r =184.113,

Q^o_m =114.285,T^o=0.368 year.

The inventory holding costs are 10,579.1. The linear costs are -8,500. The total costs are 2,079.1.

The optimal integer solution is:

R^o=2,M^o=1,Q^o_r =204.896, Q^o_m =102.448,T^o=0.41 year.

The inventory holding costs are 10,615.1. The linear costs are -8,500. The total costs are 2,115.1.

The inventory related costs of the integer solution are 0.3 per- cent higher than that of costs of continuous solution.

Example 2.

Let the parameters be in the following example:

λ=1.000 piece/year,r=0.9,u=0.48, K_m=$ 750,K_r =$ 100,

h_m=$ 200,h_r =$ 50,h_n=$ 20, c_m=$ 20,c_r =$ 15,c_d=$ -35.

The difference of these two examples are that the reuse rate is lower in the following problem. In this example the remanu- facturing firm sells the disposed items. The optimal continuous solution for this problem is:

R^o=1.489,M^o=1,Q^o_r =176.573, Q^o_m=128.467,T^o=0.368 year.

The inventory holding costs are 10,845.2. The linear costs are -10,700. The total costs are 145.2.

The optimal integer solution is:

R^o=3,M^o=2,Q_r^o=174.747, Q^o_m=126.206,T^o=0.728 year.

The inventory holding costs are 10,887.6. The linear costs are -10,700. The total costs are 187.6.

The inventory related costs of the integer solution are 0.4 per- cent higher than that of costs of continuous solution.

The solution of the border line is

R^o=2,M^o=1,Q^o_r =204.441, Q^o_m=110.739,T^o=0.426 year.

The inventory holding costs are 10,910.8. The linear costs are -10,700. The total costs are 210.8. The optimal integer solution for the total costs are 12.4 percent higher than that of of the in- teger solution on the border line. The isocost line of the optimal solution is shown on Fig. 4.

4 Calculation of the optimal reuse rateu

We determine the optimal reuse rate from the continuous cost function:

C^R(u)=















































u r

2λK_r

h_r+¹_rh_n

+(1−u)√

2λK_mh_m+ uλ (c_r−c_m−c_d)+λ (c_m+c_dr) ,

0≤u<≤u₁(r) r

2λ (K_r+K_m)h h_r+¹_rh_n

u²+h_m(1−u)²i + uλ (c_r−c_m−c_d)+λ (c_m+c_dr) ,

u₁(r)≤u≤u₂(r) u√

2λK_r(h_r+h_n)+ r

2λK_m

h(1−u)²h_m+u² 1

r −1

h_n i

+ uλ (c_r−c_m−c_d)+λ (c_m+c_dr) ,

u₁(r)≤u≤r

.

In this model we have assumed, thatr2 ≤ r. (If return rater not greater thanr2, then we have one or two interval to look for the optimal reuse rate.) It means that we have three types batch numbers. We have seen that functionC^R(u)is continuous and continuously differentiable on interval 0≤u ≤r. It is obvious